China TST 2020/4 Show that the following equation has finitely many solutions $(t,A,x,y,z)$ in positive integers $\sqrt{t(1-A^{-2})(1-x^{-2})(1-y^{-2})(1-z^{-2})}=(1+x^{-1})(1+y^{-1})(1+z^{-1})$

Bump! Any solution?

Solved with nukelauncher Rearrange the given equation to \[t\left( 1-A^{-2} \right)=\frac{(x+1)(y+1)(z+1)}{(x-1)(y-1)(z-1)},\]and define $f(x,y,z)=\frac{(x+1)(y+1)(z+1)}{(x-1)(y-1)(z-1)}$. Suppose FSoC that there were infinitely many solutions to the equation. Claim: We have $t\leq 36$. Proof. Note that $A=1$ doesn't yield any solutions, and neither does $x=1$ or $y=1$ or $z=1$.Then $1-A^{-2}\geq\frac34$ and as $f(x,y,z)$ is obviously decreasing in all of $x,y,z$ then $f(x,y,z)\leq f(2,2,2)=27$. Combining yields $t\leq 36$, as desired. $\blacksquare$ Therefore, since $t$ is bounded, there exists some $T$ such that there are infinitely many solutions to the original equation with $t=T$. Fix $t=T$. Now, since $f(x,y,z)$ is decreasing in each of $x,y,z$, this means that there are only finitely many solutions for a fixed $A$; consequently, there are solutions for infinitely many values of $A$. Claim: For any integer $N$, there are only finitely many solutions with $t=T$ and the median of $x,y,z$ at most $N$. Proof. WLOG only look at solutions with $x\geq y\geq z$. Suppose infinitely many solutions satisfy $y\leq N$ for some large integer $N$. Then there exist $y_0,z_0$ such that there are infinitely many solutions with $y=y_0$ and $z=z_0$. But now $f(x,y_0,z_0)$ is decreasing in $x$ and $T\left( 1-A^{-2} \right)$ is increasing in $A$, so there can be at most one solution, contradiction. $\blacksquare$ Thus, only finitely many solutions satisfy $y\leq N$. For all other solutions, taking $N$ large enough, we can note that \[\left\{ f(x,y,z) \right\}\leq \left\{ f(N,N,z) \right\}\leq \left\{ \left( \frac{N+1}{N-1} \right)^2\cdot \frac{z+1}{z-1} \right\}<c\]for some fixed constant $c<1$. Adding back in the finitely many solutions that were disregarded, there is a constant $C<1$ such that $\left\{ f(x,y,z) \right\}<C$ for all solutions $(T,A,x,y,z)$. However, note that there must be solutions for infinitely many $A$, and \[\left\{ T\left( 1-A^{-2} \right) \right\}=\left\{ f(x,y,z) \right\}<C\]fails for large enough $A$, contradiction. Thus, there can only be finitely many solutions to the equation, as desired.

Consider the equation $t(1-A^{-2})=\frac{(x+1)(y+1)(z+1)}{(x-1)(y-1)(z-1)}$ Claim: $t\le 36$. Proof: $\frac{3t}{4} \le 27$ WLOG, $x\le y\le z$. Claim: There exists finitely many solutions with $y\ge 22$. Proof. If $y\ge 22$, note $\frac{(y+1)(z+1)}{(y-1)(z-1)} < (\frac{23}{21})^2 < \frac 65$. If $x=2, $ we can see that for $t>6$ there is no solution and if $t\in \{4,5\}$ there are finitely many solutions (as $A$ is bounded above) If $x=3, $ again we can see $t>5$ yields no solution and $t\in \{3,4\}$ yields finitely many solutions. Otherwise, $\frac{x+1}{x-1} \le \frac 53$, so we can see that $t>3$ yields no solution and $t\in \{2,3\}$ yields finitely many solutions. Claim: For a fixed triplet $(t,x,y)$ there exist finitely many $z$. Proof. Suppose $\frac{(x-1)(y-1)t}{(x+1)(y+1)} > 1$, or otherwise no $z$ work. Notice $$\frac 34 \le \frac{(x+1)(y+1)}{(x-1)(y-1)t} \cdot \frac{z+1}{z-1}$$, which indeed shows that $z$ is bounded, as desired.